Examine the plot of the enzyme-catalyzed reaction below:
a) Is this reaction better characterized by Michaelis-Menten kinetics or simple mass action?
b) What is the approximate value of Vmax? Km?
c) Assuming that this plot was made at 100 nM enzyme, draw the curve, including axis labels, for 300 nM enzyme.
d) Imagine that this enzyme is inhibited by a competitive inhibitor with Ki (for inhibitor) = Km (for substrate). Note that v = (kcat [E] tot [S]/Km) / (1 + [S]/Km + [I]/Ki). Draw the curve of reaction velocity (v) versus [S] for [I] = 1 mM.
e) Now consider an experiment where you simultaneously introduce substrate and competitive inhibitor at equal concentrations. Make a plot of v versus concentration of [S] = [I], i.e. the X-axis should be the concentration of both I and S, which are equal.
The Correct Answer and Explanation is :
To analyze the questions posed, we first need to examine the plots from the given links. Since I cannot view the URLs directly, let’s proceed based on standard interpretation of enzyme kinetics from a typical Michaelis-Menten plot.
a) Michaelis-Menten Kinetics vs. Simple Mass Action
For an enzyme-catalyzed reaction, Michaelis-Menten kinetics is more likely if the plot of reaction velocity (v) versus substrate concentration ([S]) shows a hyperbolic curve. Simple mass action, which is more linear, typically does not apply to enzyme reactions unless they involve a single-step, non-saturating reaction. Enzyme kinetics usually display saturation at higher substrate concentrations, indicating a leveling off of the reaction velocity, which is characteristic of Michaelis-Menten kinetics.
b) Approximate values of ( V_{max} ) and ( K_m )
- ( V_{max} ): This is the maximum velocity of the reaction, observed as the plateau of the curve. To estimate ( V_{max} ), identify the highest stable velocity value on the vertical axis.
- ( K_m ): The Michaelis constant, ( K_m ), is typically estimated as the substrate concentration at which the reaction velocity is half of ( V_{max} ). It can be visually estimated from the point where the curve reaches 50% of ( V_{max} ).
c) Curve for 300 nM enzyme
If the original plot was made with 100 nM enzyme, increasing the enzyme concentration to 300 nM would generally increase the maximum velocity (( V_{max} )) proportionally, assuming other conditions remain constant. The ( K_m ) remains the same because it’s a characteristic of the enzyme-substrate interaction, not enzyme concentration. The curve would be steeper and reach a higher ( V_{max} ) more quickly.
d) Competitive inhibition curve with ( [I] = 1 mM )
With competitive inhibition, the presence of inhibitor increases the apparent ( K_m ) (denoted as ( K_m’ )) without affecting ( V_{max} ). The modified Michaelis-Menten equation in the presence of an inhibitor becomes more complex, and ( K_m’ ) increases based on the inhibitor concentration and its ( K_i ) (which equals ( K_m ) in this scenario). This shifts the curve rightwards, requiring more substrate to achieve the same velocities as without the inhibitor.
e) Plot of ( v ) versus concentration of ( [S] = [I] )
Here, the inhibitor and substrate concentrations are equal. The curve would be similar to the competitive inhibition scenario, but since both ( [S] ) and ( [I] ) increase together, the curve would continuously shift rightwards as the concentration increases, displaying a lower velocity at each substrate concentration compared to the uninhibited case.
For the visual illustrations, I will generate the plots for (c), (d), and (e) based on the provided information and typical enzyme kinetics behavior. Let’s begin with generating the curve for scenario (c).
The plot above illustrates the enzyme kinetics at two different enzyme concentrations: 100 nM (solid line) and 300 nM (dashed line). As you can see:
- The maximum velocity (( V_{max} )) of the reaction is three times higher for the 300 nM enzyme concentration compared to the 100 nM. This is because ( V_{max} ) is directly proportional to the enzyme concentration, assuming all other factors remain constant.
- The Michaelis constant (( K_m )) remains the same for both enzyme concentrations, indicating that the substrate concentration required to reach half of ( V_{max} ) is unchanged. This reinforces that ( K_m ) is a characteristic of the enzyme-substrate affinity and is not influenced by enzyme concentration.
Next, I will generate the plot for scenario (d), where the enzyme is inhibited by a competitive inhibitor with the inhibitor concentration at 1 mM, and for scenario (e), plotting ( v ) versus the concentration of ( [S] = [I] ). Let’s continue with generating those curves.
The plot illustrates the effect of competitive inhibition on enzyme kinetics:
- No inhibitor (solid line): This is the base curve with no competitive inhibitor, showing typical Michaelis-Menten kinetics.
- 1 mM inhibitor (dashed line): Here, the presence of a competitive inhibitor at a concentration of 1 mM increases the apparent ( K_m ) (denoted as ( K_m’ )), effectively requiring higher substrate concentrations to achieve the same velocities. The ( V_{max} ) remains unaffected, but the curve is shifted to the right, indicating that higher substrate concentrations are necessary to overcome the inhibition.
- [S] = [I] (dash-dot line): This scenario models equal concentrations of substrate and inhibitor. As both concentrations increase equally, the curve reflects continuous competitive inhibition, showing lower velocities at each substrate concentration compared to the uninhibited case. This illustrates the typical effect of a competitive inhibitor where it binds to the enzyme in place of the substrate, increasing the effective ( K_m ) without altering ( V_{max} ).
These plots provide a clear visual representation of how enzyme kinetics are modulated by enzyme concentration and competitive inhibition, important for understanding enzyme behavior in biochemical pathways or drug interactions.